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Steven Gubkin
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Is $\mathbb{C}^2$ homeomorphic to $\mathbb{C}^2 - (0,0)$ with the Zariski topology?

A fellow grad student asked me this, I have been playing for a while but have not come up with anything. Note that $\mathbb{C}$ is homeomorphic to $\mathbb{C} - \{0\}$ in the Zariski topology - just take any bijection and the closed sets (finite sets) will biject as well. Concocting a similar thing for the plane is harder though.

I think I can show that the rational plane and the rational plane minus the origin are homeomorphic by enumerating the irreducible curves and using a back and forth argument, but I have not written it all up formally to see if I am missing something yet.

I know the question isn't natural from the point of view of algebraic geometry, because one of the objects isn't even a variety. I think it is still interesting just to see how weird the zariski topology really is.

Steven Gubkin
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